JEE Advanced 2024 — Paper 1 — CHEM-I solution

JEE Advanced2024Paper 1CHEM-I

A closed vessel contains $10$ g of an ideal gas X at $300$ K, which exerts $2$ atm pressure. At the same temperature, $80$ g of another ideal gas Y is added to it and the pressure becomes $6$ atm. The ratio of root mean square velocities of X and Y at $300$ K is

A
$22 : 3$
B
$22 : 1$
C
$1 : 2$
D
$2 : 1$

Solution

Partial pressure of Y after addition: $P_{Y} = P_{t o t a l} - P_{X} = 6 - 2 = 4$ atm. Both gases share the same temperature and volume.

From $P V = n R T = \frac{w}{M} R T$ at fixed $V, T$ : $M \propto \frac{w}{P}$ .

$\frac{M _{X}}{M _{Y}} = \frac{w _{X} / P _{X}}{w _{Y} / P _{Y}} = \frac{10/2}{80/4} = \frac{5}{20} = \frac{1}{4}$ .

For rms speed, $v_{r m s} \propto \frac{1}{M}$ , so

$\frac{( v _{r m s} ) _{X}}{( v _{r m s} ) _{Y}} = \frac{M _{Y}}{M _{X}} = 4 = 2$ .

$∴$ The ratio is $2 : 1$ , option (D).

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